A particle is released from height S from the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the particle be at height $h$ from the surface of the Earth.By the law of conservation of mechanical energy, the total energy at height $S$ is

Vedclass · Accepted Answer

$\frac{S}{4}, \sqrt{\frac{3 gS}{2}}$

Vedclass · Answer

(A) Let the particle be at height $h$ from the surface of the Earth.By the law of conservation of mechanical energy, the total energy at height $S$ is equal to the total energy at height $h$.Total energy $E = mgS$.At height $h$, the potential energy is $PE = mgh$ and kinetic energy is $KE = mg(S-h)$.Given that $KE = 3 PE$, we have $mg(S-h) = 3 mgh$.Dividing by $mg$, we get $S-h = 3h$, which implies $S = 4h$ or $h = S/4$.Now, to find the speed $V$ at this height, we use $KE = \frac{1}{2} mV^2$.Since $KE = 3 PE = 3 mgh = 3 mg(S/4) = \frac{3 mgS}{4}$.Equating the two expressions for $KE$: $\frac{1}{2} mV^2 = \frac{3 mgS}{4}$.Solving for $V$, we get $V^2 = \frac{3 gS}{2}$, so $V = \sqrt{\frac{3 gS}{2}}$.Thus, the height is $S/4$ and the speed is $\sqrt{\frac{3 gS}{2}}$.

$A$ particle is released from height $S$ from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the Earth and the speed of the particle at that instant are respectively:

Similar Questions

The horizontal and vertical displacements of a projectile at time $t$ are $x=36 t$ and $y=48 t-4.9 t^2$,respectively. Initial velocity of the projectile in $m/s$ is

The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M \ NH_4OH$ and $25 \ mL$ of $0.2 \ M \ NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

The area (in square units) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^2-3xy+2y^2=0$ is

If $\int e^x \left( \frac{1-\sin x}{1-\cos x} \right) dx = f(x) + \text{constant}$,then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module